2017B AMC 12 真题（P版）

1 答案 E

\(m\) 个月后，Kymbrea 有 \(30 + 2m\) 本漫画书，LaShawn 有 \(10 + 6m\) 本。令 \(10 + 6m = 2(30 + 2m)\)，得到 \(10 + 6m = 60 + 4m\)，因此 \(2m = 50\)，\(m = 25\)。

所以正确答案是 E。

English

After \(m\) months, Kymbrea has \(30 + 2m\) comic books and LaShawn has \(10 + 6m.\) Setting \(10 + 6m = 2(30 + 2m)\) gives \(10 + 6m = 60 + 4m,\) so \(2m = 50\) and \(m = 25.\)

Thus, the correct answer is E.

2 答案 E

将 \(y \gt -1\) 与 \(z \gt 1\) 相加，得到 \(y + z \gt 0\)，所以 \(y + z\) 总是正数。其他四个选项都可以取负值：当 \(x = \tfrac18\)，\(y = -\tfrac14\)，\(z = \tfrac32\) 时，\(y + x^2\)，\(y + xz\)，\(y + y^2\)，和 \(y + 2y^2\) 都为负。

所以正确答案是 E。

English

Adding \(y \gt -1\) and \(z \gt 1\) gives \(y + z \gt 0,\) so \(y + z\) is always positive. Each of the other four choices can be made negative: with \(x = \tfrac18,\) \(y = -\tfrac14,\) \(z = \tfrac32,\) every one of \(y + x^2,\) \(y + xz,\) \(y + y^2,\) and \(y + 2y^2\) is negative.

Thus, the correct answer is E.

3 答案 D

原方程给出 \(3x + y = -2(x - 3y)\) \(= -2x + 6y\)，所以 \(5x = 5y\)，即 \(x = y\)。因此 \[\frac{x + 3y}{3x - y} = \frac{y + 3y}{3y - y} = \frac{4y}{2y} = 2\text{。}\]

所以正确答案是 D。

English

The equation gives \(3x + y = -2(x - 3y)\) \(= -2x + 6y,\) so \(5x = 5y,\) meaning \(x = y.\) Then \[\frac{x + 3y}{3x - y} = \frac{y + 3y}{3y - y} = \frac{4y}{2y} = 2.\]

Thus, the correct answer is D.

4 答案 C

设总距离为 \(2d\)，因此她以 \(17\) 千米/小时骑了 \(d\) 千米，又以 \(5\) 千米/小时走了 \(d\) 千米。总时间（小时）满足 \[\frac{d}{17} + \frac{d}{5} = \frac{44}{60}\text{。}\] 左边合并得到 \(\dfrac{22d}{85} = \dfrac{11}{15}\)，所以 \(d = \dfrac{11}{15} \cdot \dfrac{85}{22} = \dfrac{17}{6} = 2.833\ldots\)。她步行了约 \(2.8\) 千米。

所以正确答案是 C。

English

Let \(2d\) be the total distance, so she biked \(d\) at \(17\) km/h and walked \(d\) at \(5\) km/h. The total time in hours is \[\frac{d}{17} + \frac{d}{5} = \frac{44}{60}.\] Combining the left side gives \(\dfrac{22d}{85} = \dfrac{11}{15},\) so \(d = \dfrac{11}{15} \cdot \dfrac{85}{22} = \dfrac{17}{6} = 2.833\ldots\) She walked about \(2.8\) kilometers.

Thus, the correct answer is C.

5 答案 B

四分位距为 \(43 - 33 = 10\)，它的 \(1.5\) 倍是 \(15\)。异常值是小于 \(33 - 15 = 18\) 或大于 \(43 + 15 = 58\) 的值。只有 \(6\) 小于 \(18\)，且没有数大于 \(58\)，所以恰好有 \(1\) 个异常值。

所以正确答案是 B。

English

The interquartile range is \(43 - 33 = 10,\) so \(1.5\) times it is \(15.\) Outliers are values less than \(33 - 15 = 18\) or greater than \(43 + 15 = 58.\) Only \(6\) falls below \(18,\) and nothing exceeds \(58,\) so there is exactly \(1\) outlier.

Thus, the correct answer is B.

6 答案 D

圆心是直径的中点 \((4, 3)\)，半径为 \(\sqrt{4^2 + 3^2} = 5\)，圆的方程为 \((x - 4)^2 + (y - 3)^2 = 25\)。令 \(y = 0\)，得到 \((x - 4)^2 = 16\)，所以 \(x = 0\) 或 \(x = 8\)。与 \(x\)-轴的另一个交点位于 \(x = 8\)。

所以正确答案是 D。

English

The center is the midpoint of the diameter, \((4, 3),\) and the radius is \(\sqrt{4^2 + 3^2} = 5.\) The circle is \((x - 4)^2 + (y - 3)^2 = 25.\) Setting \(y = 0\) gives \((x - 4)^2 = 16,\) so \(x = 0\) or \(x = 8.\) The second intersection with the \(x\)-axis is at \(x = 8.\)

Thus, the correct answer is D.

7 答案 B

因为 \(\cos(\sin(x + \pi))\) \(= \cos(-\sin(x))\) \(= \cos(\sin(x))\)，所以该函数有周期 \(\pi\)。它不可能有更小的周期：\(\cos(\sin(x)) = 1\) 当且仅当 \(\sin(x) = 0\)，这只发生在 \(\pi\) 的整数倍处，所以最大值之间相隔 \(\pi\)。最小正周期为 \(\pi\)。

所以正确答案是 B。

English

Since \(\cos(\sin(x + \pi))\) \(= \cos(-\sin(x))\) \(= \cos(\sin(x)),\) the function has period \(\pi.\) It cannot be smaller: \(\cos(\sin(x)) = 1\) exactly when \(\sin(x) = 0,\) which happens only at integer multiples of \(\pi,\) so the maxima are spaced \(\pi\) apart. The least period is \(\pi.\)

Thus, the correct answer is B.

8 答案 C

设 \(x\) 和 \(y\) 分别为短边和长边，则对角线为 \(\sqrt{x^2 + y^2}\)，并且 \(\dfrac{x^2}{y^2} = \dfrac{y^2}{x^2 + y^2}\)。令 \(r = \dfrac{x^2}{y^2}\)，右边为 \(\dfrac{y^2}{x^2 + y^2} = \dfrac{1}{r + 1}\)，因此 \(r = \dfrac{1}{r+1}\)，得 \(r^2 + r - 1 = 0\)。正根为 \(r = \dfrac{\sqrt{5} - 1}{2}\)。

所以正确答案是 C。

English

Let \(x\) and \(y\) be the short and long sides, so the diagonal is \(\sqrt{x^2 + y^2}\) and \(\dfrac{x^2}{y^2} = \dfrac{y^2}{x^2 + y^2}.\) Writing \(r = \dfrac{x^2}{y^2},\) the right side is \(\dfrac{y^2}{x^2 + y^2} = \dfrac{1}{r + 1},\) so \(r = \dfrac{1}{r+1},\) giving \(r^2 + r - 1 = 0.\) The positive root is \(r = \dfrac{\sqrt{5} - 1}{2}.\)

Thus, the correct answer is C.

9 答案 A

两个圆为 \((x+10)^2 + (y+4)^2 = 169\) 和 \((x-3)^2 + (y-9)^2 = 65\)。展开并用第一个方程减去第二个方程，\(x^2\) 与 \(y^2\) 项相消，化简为 \(x + y = 3\)。任一交点都满足该方程，因此这就是经过两个交点的直线，且 \(c = 3\)。

所以正确答案是 A。

English

The circles are \((x+10)^2 + (y+4)^2 = 169\) and \((x-3)^2 + (y-9)^2 = 65.\) Expanding and subtracting the second from the first cancels the \(x^2\) and \(y^2\) terms and simplifies to \(x + y = 3.\) Any intersection point satisfies this, so it is the line through both, and \(c = 3.\)

Thus, the correct answer is A.

10 答案 D

喜欢跳舞但说自己不喜欢的学生占全体学生的 \(60\% \cdot 20\% = 12\%\)。不喜欢跳舞且如实这么说的学生占 \(40\% \cdot 90\% = 36\%\)。在所有说自己不喜欢跳舞的人中，实际喜欢跳舞的比例为 \[\frac{12}{12 + 36} = \frac{12}{48} = \frac{1}{4} = 25\%\text{。}\]

所以正确答案是 D。

English

Students who like dancing but say they dislike it make up \(60\% \cdot 20\% = 12\%\) of all students. Students who dislike dancing and say so make up \(40\% \cdot 90\% = 36\%.\) Among everyone who says they dislike dancing, the fraction who actually like it is \[\frac{12}{12 + 36} = \frac{12}{48} = \frac{1}{4} = 25\%.\]

Thus, the correct answer is D.

11 答案 B

严格递增的单调数对应 \(\{1, \ldots, 9\}\) 的非空子集，得到 \(2^9 - 1 = 511\)。严格递减的数对应 \(\{0, 1, \ldots, 9\}\) 的子集，但要排除 \(\varnothing\) 和 \(\{0\}\)（不允许前导 \(0\)），得到 \(2^{10} - 2 = 1022\)。九个一位数在两类中被重复计算，所以总数为 \(511 + 1022 - 9 = 1524\)。

所以正确答案是 B。

English

Strictly increasing monotonous numbers correspond to nonempty subsets of \(\{1, \ldots, 9\},\) giving \(2^9 - 1 = 511.\) Strictly decreasing ones correspond to subsets of \(\{0, 1, \ldots, 9\}\) other than \(\varnothing\) and \(\{0\}\) (a leading \(0\) is not allowed), giving \(2^{10} - 2 = 1022.\) The nine single-digit numbers are counted in both, so the total is \(511 + 1022 - 9 = 1524.\)

Thus, the correct answer is B.

12 答案 D

\(z^{12} = 64\) 的根位于半径 \(64^{\frac{1}{12}} = \sqrt{2}\) 的圆上，角度为 \(30^\circ\) 的整数倍。实部为正的根对应角度 \(0, \pm 30^\circ, \pm 60^\circ\)。它们的虚部相互抵消，所以和为 \[ \begin{aligned} &\sqrt{2} + 2\sqrt{2}\cos 30^\circ \\ &\quad {}+ 2\sqrt{2}\cos 60^\circ \\ &\quad {}= \sqrt{2}\bigl(1 + \sqrt{3} + 1\bigr) \\ &\quad {}= 2\sqrt{2} + \sqrt{6} \end{aligned}\text{。}\]

所以正确答案是 D。

English

The roots of \(z^{12} = 64\) lie on the circle of radius \(64^{\frac{1}{12}} = \sqrt{2},\) at angles that are multiples of \(30^\circ.\) Those with positive real part are at angles \(0, \pm 30^\circ, \pm 60^\circ.\) Their imaginary parts cancel, so the sum is \[ \begin{aligned} &\sqrt{2} + 2\sqrt{2}\cos 30^\circ \\ &\quad {}+ 2\sqrt{2}\cos 60^\circ \\ &\quad {}= \sqrt{2}\bigl(1 + \sqrt{3} + 1\bigr) \\ &\quad {}= 2\sqrt{2} + \sqrt{6}. \end{aligned} \]

Thus, the correct answer is D.

13 答案 D

在考虑对称性之前，共有 \(\dfrac{6!}{3!2!}=60\) 种涂色。两个非恒等旋转把这些圆盘分成两个 \(3\)-循环，所以它们都不能固定颜色数量为 \(3,2,1\) 的涂色。

\(3\) 个反射中的每一个都固定 \(2\) 个圆盘，并把其余 \(4\) 个圆盘两两交换成 \(2\) 对。要使一种涂色保持不变，唯一的绿色圆盘和 \(3\) 个蓝色圆盘之一必须占据这两个不动位置，共有 \(2\) 种顺序；两组交换对中任意一组都可以是红色对，因此每个反射固定 \(2\cdot2=4\) 种涂色。于是由 Burnside 引理得到 \(\dfrac{60+3(4)}{6}=12\) 种不同的涂色。

所以正确答案是 D。

English

Before accounting for symmetry, there are \(\dfrac{6!}{3!2!}=60\) paintings. The two nonidentity rotations partition the disks into two \(3\)-cycles, so neither can fix a painting having color counts \(3,2,1.\)

Each of the \(3\) reflections fixes \(2\) disks and swaps the other \(4\) in \(2\) pairs. For a painting to be fixed, the lone green disk and one of the \(3\) blue disks must occupy the two fixed positions, in \(2\) orders. Of the two swapped pairs, either one can be the red pair, giving \(2\cdot2=4\) fixed paintings per reflection. Burnside’s Lemma therefore gives \(\dfrac{60+3(4)}{6}=12\) distinct paintings.

Thus, the correct answer is D.

14 答案 E

将圆锥台的侧边延长至一点，由相似三角形可知，该圆锥台等于一个半径为 \(2\)、高为 \(8\) 的圆锥减去一个半径为 \(1\)、高为 \(4\) 的圆锥。因此圆锥台的体积为 \[ \begin{aligned} &\tfrac13 \pi (2^2)(8) \\ &\quad {}- \tfrac13 \pi (1^2)(4) \\ &\quad {}= \tfrac{32}{3}\pi - \tfrac{4}{3}\pi \\ &\quad {}= \tfrac{28}{3}\pi\text{。} \end{aligned} \] 顶部半径为 \(2\)、高为 \(4\) 的圆锥体积为 \(\tfrac13 \pi (2^2)(4) = \tfrac{16}{3}\pi\)。总体积为 \(\tfrac{28}{3}\pi + \tfrac{16}{3}\pi = \tfrac{44}{3}\pi\)。

所以正确答案是 E。

English

Extending the frustum’s sides to a point, similar triangles show the frustum equals a cone of radius \(2\) and height \(8\) minus a cone of radius \(1\) and height \(4:\) \[ \begin{aligned} &\tfrac13 \pi (2^2)(8) \\ &\quad {}- \tfrac13 \pi (1^2)(4) \\ &\quad {}= \tfrac{32}{3}\pi - \tfrac{4}{3}\pi \\ &\quad {}= \tfrac{28}{3}\pi. \end{aligned} \] The top cone of radius \(2\) and height \(4\) adds \(\tfrac13 \pi (2^2)(4) = \tfrac{16}{3}\pi.\) The total is \(\tfrac{28}{3}\pi + \tfrac{16}{3}\pi = \tfrac{44}{3}\pi.\)

Thus, the correct answer is E.

15 答案 E

令 \(X = [\triangle ABC]\)，并作线段 \(CB'\)、\(AC'\) 和 \(BA'\)。三角形 \(BB'C\) 的底边 \(BB' = 3 \cdot AB\)，且从 \(C\) 到直线 \(AB\) 的高与 \(\triangle ABC\) 相同，所以面积为 \(3X\)；同理，\(\triangle CC'A\) 和 \(\triangle AA'B\) 的面积也各为 \(3X\)。接着，\(\triangle AA'C'\) 与 \(\triangle ACC'\) 高相同，而前者的底边为后者的 \(3\) 倍，所以面积为 \(9X\)；同理，\(\triangle CC'B'\) 和 \(\triangle BB'A'\) 的面积也各为 \(9X\)。因此 \[ \begin{aligned} &[\triangle A'B'C'] = X \\ &\quad {}+ 3(3X) + 3(9X) \\ &\quad {}= 37X\text{，} \end{aligned} \] 所以面积比为 \(37 : 1\)。

所以正确答案是 E。

English

Let \(X = [\triangle ABC],\) and draw segments \(CB',\) \(AC',\) and \(BA'.\) Triangle \(BB'C\) has base \(BB' = 3 \cdot AB\) and the same altitude as \(\triangle ABC\) from \(C\) to line \(AB,\) so its area is \(3X;\) likewise \(\triangle CC'A\) and \(\triangle AA'B\) each have area \(3X.\) Next, \(\triangle AA'C'\) has \(3\) times the base and the same height as \(\triangle ACC',\) so its area is \(9X;\) similarly \(\triangle CC'B'\) and \(\triangle BB'A'\) each have area \(9X.\) Thus \[ \begin{aligned} &[\triangle A'B'C'] = X \\ &\quad {}+ 3(3X) + 3(9X) \\ &\quad {}= 37X, \end{aligned} \] so the ratio is \(37 : 1.\)

Thus, the correct answer is E.

16 答案 B

\(21!\) 中 \(2\) 的指数为 \(\lfloor \frac{21}{2} \rfloor\) \(+ \lfloor \frac{21}{4} \rfloor\) \(+ \lfloor \frac{21}{8} \rfloor\) \(+ \lfloor \frac{21}{16} \rfloor\) \(= 10 + 5 + 2 + 1\) \(= 18\)。每个因数都形如 \(2^i b\)，其中 \(0 \le i \le 18\) 且 \(b\) 为奇数；它为奇数正好在 \(i = 0\) 时。因此奇因数的比例为 \(\dfrac{1}{18 + 1} = \dfrac{1}{19}\)。

所以正确答案是 B。

English

The exponent of \(2\) in \(21!\) is \(\lfloor \frac{21}{2} \rfloor\) \(+ \lfloor \frac{21}{4} \rfloor\) \(+ \lfloor \frac{21}{8} \rfloor\) \(+ \lfloor \frac{21}{16} \rfloor\) \(= 10 + 5 + 2 + 1\) \(= 18.\) Every divisor has the form \(2^i b\) with \(0 \le i \le 18\) and \(b\) odd; it is odd exactly when \(i = 0.\) So the fraction of odd divisors is \(\dfrac{1}{18 + 1} = \dfrac{1}{19}.\)

Thus, the correct answer is B.

17 答案 D

令 \(p = \dfrac23\)。游戏 A 在三次抛掷结果全部相同时获胜，概率为 \(p^3 + (1-p)^3\)。游戏 B 要求第一对相同且第二对相同，每对相同的概率为 \(p^2 + (1-p)^2\)，所以获胜概率为 \(\bigl(p^2 + (1-p)^2\bigr)^2\)。当 \(p = \tfrac23\) 时，游戏 A 的概率为 \(\left(\tfrac23\right)^3 + \left(\tfrac13\right)^3 = \tfrac{9}{27} = \tfrac13\)，游戏 B 的概率为 \(\left(\tfrac49 + \tfrac19\right)^2 = \left(\tfrac59\right)^2 = \tfrac{25}{81}\)。差为 \(\tfrac{27}{81} - \tfrac{25}{81} = \tfrac{2}{81}\)，所以游戏 A 的胜率高 \(\tfrac{2}{81}\)。

所以正确答案是 D。

English

Let \(p = \dfrac23.\) Game A is won when all three tosses match: \(p^3 + (1-p)^3.\) Game B needs the first pair to match and the second pair to match, each with probability \(p^2 + (1-p)^2,\) so the win probability is \(\bigl(p^2 + (1-p)^2\bigr)^2.\) With \(p = \tfrac23,\) Game A gives \(\left(\tfrac23\right)^3 + \left(\tfrac13\right)^3 = \tfrac{9}{27} = \tfrac13,\) and Game B gives \(\left(\tfrac49 + \tfrac19\right)^2 = \left(\tfrac59\right)^2 = \tfrac{25}{81}.\) The difference is \(\tfrac{27}{81} - \tfrac{25}{81} = \tfrac{2}{81},\) so Game A is \(\tfrac{2}{81}\) more likely.

Thus, the correct answer is D.

18 答案 D

因为 \(\angle ACB\) 是半圆上的圆周角，所以它是直角，因此 \(\triangle ABC \sim \triangle AED\)（都是直角三角形且共享角 \(A\)）。它们的面积比为 \(AB^2 : AE^2\)。这里 \(AB = 4\)，所以 \(AB^2 = 16\)，且 \(AD = AB + BD = 7\)，因此 \(AE^2 = AD^2 + ED^2\) \(= 49 + 25\) \(= 74\)。\(\triangle AED\) 的面积为 \(\tfrac12 \cdot 7 \cdot 5 = \tfrac{35}{2}\)。所以 \[[\triangle ABC] = \frac{16}{74} \cdot \frac{35}{2} = \frac{140}{37}\text{。}\]

所以正确答案是 D。

English

Since \(\angle ACB\) is inscribed in a semicircle, it is a right angle, so \(\triangle ABC \sim \triangle AED\) (both right-angled and sharing angle \(A\)). Their areas are in ratio \(AB^2 : AE^2.\) Here \(AB = 4,\) so \(AB^2 = 16,\) and \(AD = AB + BD = 7,\) so \(AE^2 = AD^2 + ED^2\) \(= 49 + 25\) \(= 74.\) The area of \(\triangle AED\) is \(\tfrac12 \cdot 7 \cdot 5 = \tfrac{35}{2}.\) Thus \[[\triangle ABC] = \frac{16}{74} \cdot \frac{35}{2} = \frac{140}{37}.\]

Thus, the correct answer is D.

19 答案 C

\(N\) 的最后一位是 \(4\)，所以 \(N \equiv 4 \pmod 5\)。对模 \(9\)，求数字和：数字 \(1\) 到 \(9\) 贡献它们自身的数字，\(10\) 到 \(44\) 的十位数字与个位数字合计为 \(270\)，这是 \(9\) 的倍数，所以 \(N \equiv 0 \pmod 9\)。因此 \(N - 9\) 是 \(9\) 的倍数，且它的最后一位是 \(5\)，所以也是 \(5\) 的倍数；于是 \(N - 9\) 是 \(45\) 的倍数。因此 \(N \equiv 9 \pmod{45}\)。

所以正确答案是 C。

English

The last digit of \(N\) is \(4,\) so \(N \equiv 4 \pmod 5.\) For mod \(9,\) sum the digits: the numbers \(1\)–\(9\) contribute their digits, the tens digits of \(10\)–\(44\) and the units digits together sum to \(270,\) which is a multiple of \(9,\) so \(N \equiv 0 \pmod 9.\) The number \(N - 9\) is then a multiple of \(9,\) and its last digit is \(5,\) so it is a multiple of \(5;\) hence \(N - 9\) is a multiple of \(45.\) Therefore \(N \equiv 9 \pmod{45}.\)

Thus, the correct answer is C.

20 答案 D

对每个正整数 \(n\)，\(\lfloor \log_2 x \rfloor = -n\) 当且仅当 \(\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}}\)，这是长度为 \(\dfrac{1}{2^n}\) 的区间。两个取整值都等于 \(-n\) 的事件是面积为 \(\dfrac{1}{4^n}\) 的正方形。对所有 \(n\) 求和，概率为 \[\sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}\text{。}\]

所以正确答案是 D。

English

For each positive integer \(n,\) \(\lfloor \log_2 x \rfloor = -n\) exactly when \(\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}},\) an interval of length \(\dfrac{1}{2^n}.\) The event that both floors equal \(-n\) is a square of area \(\dfrac{1}{4^n}.\) Summing over all \(n,\) the probability is \[\sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}.\]

Thus, the correct answer is D.

21 答案 E

令 \(S\) 为七个分数之和。则 \(S\) 是 \(7\) 的倍数且 \(658 \le S \le 679\)，所以 \(S \in \{658, 665, 672, 679\}\)。因为前六次测验后的平均数是整数，\(S - 95\) 必须是 \(6\) 的倍数，这迫使 \(S = 665\)。因此前六个分数之和为 \(570\)，它是 \(5\) 的倍数。前五次测验后的平均数也是整数，所以前五个分数之和是 \(5\) 的倍数，于是第六个分数也是 \(5\) 的倍数。由于所有分数互不相同且第七个分数为 \(95\)，第六个分数必须是 \(100\)。

所以正确答案是 E。

English

Let \(S\) be the sum of all seven scores. Then \(S\) is a multiple of \(7\) with \(658 \le S \le 679,\) so \(S \in \{658, 665, 672, 679\}.\) Since the average after six tests is an integer, \(S - 95\) is a multiple of \(6,\) which forces \(S = 665.\) Then the first six scores sum to \(570,\) a multiple of \(5;\) the average after five tests is an integer, so the first five scores also sum to a multiple of \(5,\) making the sixth score a multiple of \(5.\) Since all scores differ and the seventh is \(95,\) the sixth must be \(100.\)

Thus, the correct answer is E.

22 答案 B

每轮有 \(4 \cdot 3 = 12\) 个等可能的（给出者，接收者）有序对，所以共有 \(12^4\) 个结果序列。每个人最后都有四枚硬币，当且仅当四次转移相互抵消。有利模式为：一次 \(4\)-循环赠送（\(24 \cdot 6 = 144\) 种），两组不相交的相互交换（\(24 \cdot 3 = 72\) 种），同一对玩家互相交换两次（\(6 \cdot 6 = 36\) 种），以及某个玩家既给另外两人各一枚又从这两人各收一枚（\(4 \cdot 3 \cdot 24 = 288\) 种）。总数为 \(144 + 72 + 36 + 288 = 540\)，概率为 \(\dfrac{540}{12^4} = \dfrac{540}{20736} = \dfrac{5}{192}\)。

所以正确答案是 B。

English

Each round has \(4 \cdot 3 = 12\) equally likely (giver, receiver) pairs, so there are \(12^4\) outcome sequences. Everyone ends with four coins exactly when the four transfers cancel. The favorable patterns are: a \(4\)-cycle of gifts (\(24 \cdot 6 = 144\) ways), two disjoint mutual exchanges (\(24 \cdot 3 = 72\)), one pair exchanging twice (\(6 \cdot 6 = 36\)), and one player both giving to and receiving from each of two others (\(4 \cdot 3 \cdot 24 = 288\)). These total \(144 + 72 + 36 + 288 = 540.\) The probability is \(\dfrac{540}{12^4} = \dfrac{540}{20736} = \dfrac{5}{192}.\)

Thus, the correct answer is B.

23 答案 D

点 \(A, B, C\) 都在 \(y = x^2\) 上，所以 \(g(x) = f(x) - x^2\) 以 \(2, 3, 4\) 为根，即 \(g(x) = a(x-2)(x-3)(x-4)\)，其中 \(a \ne 0\)。\(x^3\) 和 \(x^2\) 在 \(f\) 中的系数分别为 \(a\) 和 \(1 - 9a\)，因此由韦达定理，\(f(x) - L(x)\)（对任意一次函数 \(L\)）的三个根之和为 \(9 - \tfrac1a\)。直线 \(AB, AC, BC\) 与三次曲线相交的三个横坐标集合分别为 \(\{2, 3, x_D\}\)、\(\{2, 4, x_E\}\)、\(\{3, 4, x_F\}\)，所以 \[ \begin{aligned} &x_D + x_E + x_F \\ &\quad {}= 3\left(9 - \tfrac1a\right) \\ &\quad {}- 2(2 + 3 + 4) \\ &\quad {}= 9 - \tfrac3a = 24\text{，} \end{aligned} \] 解得 \(a = -\tfrac15\)。因此 \(f(x) = x^2\) \(- \tfrac15(x-2)(x-3)(x-4)\)，所以 \(f(0) = 0 - \tfrac15(-2)(-3)(-4)\) \(= \tfrac{24}{5}\)。

所以正确答案是 D。

English

The points \(A, B, C\) lie on \(y = x^2,\) so \(g(x) = f(x) - x^2\) has roots \(2, 3, 4:\) \(g(x) = a(x-2)(x-3)(x-4)\) for some \(a \ne 0.\) The coefficients of \(x^3\) and \(x^2\) in \(f\) are \(a\) and \(1 - 9a,\) so by Vieta the three roots of \(f(x) - L(x)\) (for any linear \(L\)) sum to \(9 - \tfrac1a.\) The lines \(AB, AC, BC\) meet the cubic in triples \(\{2, 3, x_D\},\) \(\{2, 4, x_E\},\) \(\{3, 4, x_F\},\) so \[ \begin{aligned} &x_D + x_E + x_F \\ &\quad {}= 3\left(9 - \tfrac1a\right) \\ &\quad {}- 2(2 + 3 + 4) \\ &\quad {}= 9 - \tfrac3a = 24, \end{aligned} \] giving \(a = -\tfrac15.\) Then \(f(x) = x^2\) \(- \tfrac15(x-2)(x-3)(x-4),\) so \(f(0) = 0 - \tfrac15(-2)(-3)(-4)\) \(= \tfrac{24}{5}.\)

Thus, the correct answer is D.

24 答案 D

令 \(BC = 1\)，\(AB = r \gt 1\)。由 \(\triangle ABC \sim \triangle BCD\) 以及直角位置，可取 \(C = (0,0)\)，\(B = (0,1)\)，\(A = (r,1)\)，\(D = \bigl(\tfrac1r, 0\bigr)\)。设 \(E = (x, y)\)，其中 \(x, y \gt 0\)。由 \(\triangle ABC \sim \triangle CEB\) 得 \(\dfrac{x}{y} = \tan(\angle ECB)\) \(= \tan(\angle BAC)\) \(= \dfrac1r\) 且 \(x^2 + y^2 = \dfrac{r^2}{1 + r^2}\)，所以 \(x = \dfrac{r}{1+r^2}\)，\(y = \dfrac{r^2}{1+r^2}\)。这两个相关的面积为 \[ \begin{aligned} [\triangle CEB]&=\dfrac{r}{2(1+r^2)},\\ [\triangle AED]&=\dfrac{r^4-r^2+1}{2r(1+r^2)} \end{aligned}\text{。}\] 令后者等于前者的 \(17\) 倍，得到 \(r^4 - 18r^2 + 1 = 0\)。于是 \(r^2 = 9 + 4\sqrt5 = (2 + \sqrt5)^2\)，所以 \(r = 2 + \sqrt5\)。

所以正确答案是 D。

English

Set \(BC = 1\) and \(AB = r \gt 1.\) The similarity \(\triangle ABC \sim \triangle BCD\) with the right angles places the figure at \(C = (0,0),\) \(B = (0,1),\) \(A = (r,1),\) \(D = \bigl(\tfrac1r, 0\bigr).\) Let \(E = (x, y)\) with \(x, y \gt 0.\) From \(\triangle ABC \sim \triangle CEB\) we get \(\dfrac{x}{y} = \tan(\angle ECB)\) \(= \tan(\angle BAC)\) \(= \dfrac1r\) and \(x^2 + y^2 = \dfrac{r^2}{1 + r^2},\) so \(x = \dfrac{r}{1+r^2},\) \(y = \dfrac{r^2}{1+r^2}.\) The two relevant areas are \[ \begin{aligned} [\triangle CEB]&=\dfrac{r}{2(1+r^2)},\\ [\triangle AED]&=\dfrac{r^4-r^2+1}{2r(1+r^2)}. \end{aligned} \] Setting the second equal to \(17\) times the first gives \(r^4 - 18r^2 + 1 = 0.\) Then \(r^2 = 9 + 4\sqrt5 = (2 + \sqrt5)^2,\) so \(r = 2 + \sqrt5.\)

Thus, the correct answer is D.

25 答案 D

设球队数为 \(T\)。对 \(9\) 人子集求和时，每支球队被计数 \(\binom{n-5}{4}\) 次；对 \(8\) 人子集求和时，每支球队被计数 \(\binom{n-5}{3}\) 次。两个平均值分别为 \(\dfrac{\binom{n-5}{4}T}{\binom n9}\) 和 \(\dfrac{\binom{n-5}{3}T}{\binom n8}\)。令第一个等于第二个的倒数并化简，得到 \[ \begin{aligned} &T \\ &\quad {}= \scriptsize \frac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7} \end{aligned}\text{。}\] 我们需要它在 \(n \ge 9\) 时为正整数。令 \(N =\) \(n(n-1)(n-2)(n-3)(n-4)\)；作为五个连续整数的乘积，\(N\) 总能被 \(5\) 整除。\(N\) 是 \(7\) 的倍数这一条件对模 \(7\) 的 \(5\) 个余数成立；是 \(9\) 的倍数对模 \(9\) 的 \(7\) 个余数成立；是 \(32\) 的倍数对模 \(16\) 的 \(8\) 个余数成立。因此由中国剩余定理，模 \(\operatorname{lcm}(7,9,16)=1008\) 共有 \(5\cdot7\cdot8=280\) 个解。于是 \(1 \le n \le 2016\) 中有 \(560\) 个值；去掉 \(n = 1, 2, 3, 4\)（它们小于 \(9\)），再加上 \(n = 2017\)（因为 \(2017 \equiv 1 \pmod{1008}\)），得到 \(560 - 4 + 1 = 557\) 个有效值。

所以正确答案是 D。

English

Let \(T\) be the number of teams. Summing over size-\(9\) subsets counts each team \(\binom{n-5}{4}\) times and over size-\(8\) subsets \(\binom{n-5}{3}\) times. The averages are \(\dfrac{\binom{n-5}{4}T}{\binom n9}\) and \(\dfrac{\binom{n-5}{3}T}{\binom n8};\) setting the first equal to the reciprocal of the second and simplifying gives \[ \begin{aligned} &T \\ &\quad {}= \scriptsize \frac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7}. \end{aligned} \] We need this to be a positive integer with \(n \ge 9.\) Let \(N =\) \(n(n-1)(n-2)(n-3)(n-4);\) as a product of five consecutive integers, \(N\) is always divisible by \(5.\) The condition that \(N\) is divisible by \(7\) holds for \(5\) residues modulo \(7;\) divisibility by \(9\) holds for \(7\) residues modulo \(9;\) and divisibility by \(32\) holds for \(8\) residues modulo \(16.\) The Chinese Remainder Theorem therefore gives \(5\cdot7\cdot8=280\) solutions modulo \(\operatorname{lcm}(7,9,16)=1008.\) So there are \(560\) values in \(1 \le n \le 2016;\) removing \(n = 1, 2, 3, 4\) (which are below \(9\)) and adding \(n = 2017\) (since \(2017 \equiv 1 \pmod{1008}\)) gives \(560 - 4 + 1 = 557\) valid values.

Thus, the correct answer is D.